Question: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $y \neq 0$. $p = \dfrac{-y - 9}{-3y + 9} \times \dfrac{-4y + 12}{y^2 - y - 90} $
Answer: First factor the quadratic. $p = \dfrac{-y - 9}{-3y + 9} \times \dfrac{-4y + 12}{(y + 9)(y - 10)} $ Then factor out any other terms. $p = \dfrac{-(y + 9)}{-3(y - 3)} \times \dfrac{-4(y - 3)}{(y + 9)(y - 10)} $ Then multiply the two numerators and multiply the two denominators. $p = \dfrac{ -(y + 9) \times -4(y - 3) } { -3(y - 3) \times (y + 9)(y - 10) } $ $p = \dfrac{ 4(y + 9)(y - 3)}{ -3(y - 3)(y + 9)(y - 10)} $ Notice that $(y - 3)$ and $(y + 9)$ appear in both the numerator and denominator so we can cancel them. $p = \dfrac{ 4\cancel{(y + 9)}(y - 3)}{ -3(y - 3)\cancel{(y + 9)}(y - 10)} $ We are dividing by $y + 9$ , so $y + 9 \neq 0$ Therefore, $y \neq -9$ $p = \dfrac{ 4\cancel{(y + 9)}\cancel{(y - 3)}}{ -3\cancel{(y - 3)}\cancel{(y + 9)}(y - 10)} $ We are dividing by $y - 3$ , so $y - 3 \neq 0$ Therefore, $y \neq 3$ $p = \dfrac{4}{-3(y - 10)} $ $p = \dfrac{-4}{3(y - 10)} ; \space y \neq -9 ; \space y \neq 3 $